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1

Your reasoning is correct, for a continuous (bounded) everywhere defined operator $A$ on a Banach (in particular on a Hilbert) space, the denseness of the range of $\lambda\mathbb{I} - A$ together with the boundedness of the inverse already implies the surjectivity of $\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting ...


1

A closed densely-defined linear operator on a Hilbert space can have empty spectrum. For example, let $H=L^{2}[0,1]$ and let $A=\frac{d}{dx}$ on the domain consisting of absolutely continuous $f \in L^{2}$ for which $f(0)=0$ and $f' \in L^{2}$. To show that $A$ has no spectrum, it is enough to prove that the resolvent $R(\lambda)=(A-\lambda I)^{-1}$ exists ...


6

$$ A(A + I)^{-1} = (A + I - I)(A + I)^{-1}\\ = (A + I)(A + I)^{-1} - I(A + I)^{-1}\\ = (A + I)^{-1}(A+I) - (A + I)^{-1}I\\ = (A + I)^{-1}(A+I-I)\\ = (A + I)^{-1}A $$ since $(A + I)$ commutes with its own inverse, and $I$ commutes with anything.


1

This is only a check! The spectrum is neglible: $$0\leq\lambda_\mathbb{C}(\sigma(H))\leq\lambda_\mathbb{C}(\mathbb{R})=0$$ By functional calculus: ...


0

Since the codimension of $R$ is one, the quotient space $Y/R$ has dimension one. (For quotients of Banach spaces, see this earlier MSE question, for example.) Since $Y/R$ has dimension one, there is a linear isomorphism $f:Y/R\to\mathbb R$ (assuming you are working over the reals). The quotient map $p:Y\to Y/R$ is linear and continuous and its kernel is $R$. ...


1

Yes, since $B(H)$ is a Banach algebra it makes perfect sense to talk about questions of convergence. Unfortunately, the sum in question need not converge. Let $v$ be some unit vector in $H$. We can construct a counterexample as follows: Let $P_v$ denote the orthogonal projection onto the subspace spanned by $v$. Let $a = - P_v$. Then $a^{n} = (-1)^{n} P_v$, ...


0

copper.hat's answer is spectacular. John Ma's observation was also dead on, and provides a slick answer that I would like to elaborate on. (I suspect that at their root both answers are equivalent.) Since we are in a Hilbert space $\mathcal{H}$, every linear functional is of the form $\langle \cdot, x \rangle$ for some $x \in \mathcal{H}$. The ...


2

If I understood you well, your $\ell^0$ is the space of complex sequences that are eventually vanishing. If that is right, consider the sequence $(c_n) \in (\ell^0)^{\mathbb N}$ defined by $c_n(k) = \frac{1}{k}$ for $1 \le k \le n$ and $c_n(k) = 0$ for $k > n$. $(c_n)$ is a Cauchy sequence as for $n < m$ $$\Vert c_n - c_m \Vert^ 2=\sum_{k=n+1}^m ...


0

It's hard to say whether your expression is valid. However, it is certainly unclear since you include more than one $\mid$ in your set definition. I would write something like this: $$ \newcommand{\ip}[1]{\langle#1 \rangle} (U^\perp)^\perp = \{w \in V: v \in U^\perp \implies \ip{v,w} = 0\}\\ = \{w \in V: (\forall u \in U, \ip{v,u} = 0) \implies \ip{v,w} = ...


2

Suppose $\mathcal{H}=L^{2}[0,\pi]$. Suppose $\mathcal{M}$ is a countably dense subset of compactly supported $C^{\infty}$ functions on $(0,\pi)$. Apply Gram-Schmidt in order to obtain an orthonormal basis $\{ f_{k} \}_{k=1}^{\infty}$ of smooth compactly supported functions on $(0,\pi)$. The following operators $L_{\alpha,\beta}$ are selfadjoint $$ ...


0

Suppose that the range $U(\mathcal{H} )$ is closed. Since $U$ is one to one and continuous operator then by Banach Theorem $U$ has continuous inverse $U^{-1} :U(\mathcal{H} ) \to \mathcal{H} .$ and thus there exists a positive real number $ \sigma$ such that $$||U(t) ||\geqslant \sigma $$ for all $t\in \mathcal{H} , ||t||=1.$ Now costder a sequence $$x_n ...


2

I think you misunderstood it: the "set theoretic complement" is $H \setminus E(H)$, not $E^{\dagger}(H)$


0

There is a bound on the norm of partial trace of operator that depends on the type of norm and the size of the Hilbert space you traced out. For the Schatten $p$-norm, which is defined for $p\ge1$ as $$\|A\|_p=\left[\text{Tr}(|A|^p)\right]^{1/p}$$ one can show a bound, due to Rastegin, where $$\|\text{Tr}_1(A)\|_p \le [\dim(H_1)]^{(p-1)/p} \|A\|_p$$ This ...


1

With the constraint that the cardinality $\kappa$ of a Hamel basis shall satisfy $\kappa^{\aleph_0} = \kappa$, the answer is affirmative. For an infinite set $S$, let us denote the family of countably infinite subsets of $S$ by $\mathfrak{P}_\omega(S)$. Then we look at the space $$\ell^2(S) = \biggl\{ f\colon S \to \mathbb{K} : \sum_{s\in S} \lvert ...


0

Adjoint Formula It holds the relations: $$\mathrm{ad}_\varepsilon(A)=\delta\tau_{+\varepsilon}[A]e^{i\varepsilon H}=e^{i\varepsilon H}\delta\tau_{-\varepsilon}[A]$$ They are derivations: $$\mathrm{ad}_\varepsilon(AB)=\mathrm{ad}_\varepsilon(A)B+A\mathrm{ad}_\varepsilon(B)$$ And they vanish on: $$\mathrm{ad}_\varepsilon(e^{itH})=i[\delta ...


1

Suppose $T$ is selfadjoint with $\|T\| =1$. Then $|(Tx,x)| \le \|T\|\|x\|=1$ for all unit vectors $x$. If $|(Tx,x)|=1$ for some unit vector $x$, then the above gives $$ 1=|(Tx,x)|\le \|Tx\|\|x\|\le \|T\|\|x\|^{2}\le 1 $$ Therefore, $|(Tx,x)|=\|Tx\|\|x\|$ with $x\ne 0$ which, by Cauchy-Schwarz, forces $Tx=\alpha x$ for some scalar $\alpha$ and unit vector ...


1

You have trivially that $$C^{\infty}_c(\mathbb{R^+}) \subset C^{\infty}_0(\mathbb{R^+})$$ But $W^{1,2}_0(\mathbb{R^+})$ is usually defined as the closure of $C^{\infty}_c(\mathbb{R^+})$ for the $W^{1,2}$ norm $$W^{1,2}_0(\mathbb{R^+}) := \overline{C^{\infty}_c(\mathbb{R^+})} $$


1

Use the Cauchy criterion. If $m<n$ then $$ \Bigl\|\sum_{k=m}^na_k\,h_k\Bigr\|^2=\sum_{k=m}^n|a_k|^2, $$ which can be made smaller than any $\epsilon>0$ because $\sum_{k=1}^\infty|a_k|^2<\infty$. Since a Hilbert space is complete, this implies that $\sum_{k=1}^\infty a_k\,h_k$ is convergent.


3

Let $\kappa$ be any cardinal. Define $\ell^2(\kappa)$ as follows: As a set, we have $$ \ell^2(\kappa) := \left\{x \colon \kappa \to \mathbf K \Biggm| \sum_{i \in \kappa} |x(i)|^2 < \infty \right\} $$ With the scalar product $$ (x,y) := \sum_{i \in \kappa} x(i)\overline{y(i)} $$ $\ell^2(\kappa)$ is a Hilbert space, of which the maps $e^i$, $i \in \kappa$, ...


0

For the other part: $x=\sum a_ne_n$ and thus $<x,y_k>=<\sum a_ne_n,y_k>=\sum a_n<e_n,y_k>$


0

I'll give you one example: you can take $$ y_k=k\,e_k. $$ Then $(e_n,y_k)=0$ for all $k>n$. And if $x=\sum_n\frac1n \,e_n$, then $$ (x,y_k)=1,\ \ \text{ for all }k. $$


0

Let $D \subset H$ be dense subspace and assume $T: D \to H$ is bounded on $D$ by say $M \in \mathbb R^+$. This means $$ \sup_{ x \in D \setminus \{0\}} \frac{\Vert Tx \Vert} {\Vert x \Vert } = M. $$ We claim that there exists a linear map $\tilde T: H \to H$ which extends $T$ and has norm $M$. Now, let $y \in H$ be arbitrary. Approximate $y$ by elements ...


0

Construction Given the Hilbert space $\ell^2(\Lambda)$. Construct the spectral measure: $$E(A)\varphi:=\sum_{\lambda\in\Lambda}\delta_{\lambda\in A}\langle\sigma_\lambda,\varphi\rangle\sigma_\lambda\quad(\sigma_\lambda:=\chi_{\{\lambda\}})$$ Then one obtains: $$\sigma_0(N)=\Lambda\quad\sigma(N)=\overline{\Lambda}$$ Concluding the construction. Reference ...


1

@AdamHughes answer works just fine, I am giving here a complete proof of the next proposition: If $(X, \|\cdot \|$) is a normed space, then any finite dimensional subset is closed. Here is the proof Proof Let $M \subset X$ have dimension $k \in \mathbb{N}$, then there exist $\{ e_1, \cdots, e_k \}$ such that $$ M= \text{span}\{ e_1, \cdots, e_k \} $$ Take ...


0

Domain For the domain: $$P_\lambda N\subseteq NP_\lambda\implies P_\lambda\mathcal{D}\subseteq\mathcal{D}$$ So one obtains $$\varphi\in\mathcal{D}:\quad\varphi=\sum_\lambda P_\lambda\varphi\in\sum_\lambda\mathcal{S}_\lambda\cap\mathcal{D}$$ (That allows to continue!) Operator Regard index sets: $$\#\Lambda<\infty:\quad\lambda_0\in\Lambda$$ Then one ...


1

Since $M\cong \Bbb F^n$, $M$ is locally compact, hence closed.


1

Embedding Consider the embeddings: $$J_\alpha:\mathcal{S}_\alpha\to\mathcal{H}:\quad J_\beta^*J_\alpha=\delta_{\beta\alpha}1_\alpha\quad J_\alpha J_\alpha^*=P_\alpha$$ Construct the unitary map: $$U\varphi:=(J_\alpha^*\varphi)_\alpha\quad V(\varphi_\alpha)_\alpha:=\sum_\alpha J_\alpha\varphi_\alpha$$ Indeed they are inverses:* $$VU\varphi=\sum_\alpha ...


0

Additivity Remind on summability: $$a_{\kappa\lambda}\geq0:\quad\sum_\kappa\sum_\lambda a_{\kappa\lambda}=\sum_\lambda\sum_\kappa a_{\kappa\lambda}$$ So the measure is additive: $$\sum_k\mu(A_k)=\sum_k\sum_\lambda\mu_\lambda(\iota_\lambda^{-1}A_k)\\ ...


1

Unitary Map Construct the unitary map: $$V:\mathcal{L}^2(\nu_0)\to\mathcal{H}:\quad Vh:=h(N)\varphi_0$$ By measurable calculus: $$h\in\mathcal{L}^2(\nu_0):\quad\|h(N)\varphi_0\|^2=\int|h|^2\mathrm{d}\nu_0$$ Especially one has: $$V\chi_A=E(A)\varphi_0:\quad\mathcal{H}=\overline{\{V\chi_A:A\in\mathcal{B}(\mathbb{C})\}}$$ Concluding unitarity. ...


1

Consider the POSET: $$\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}=\mathcal{S}_\varphi$$ $$\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}\perp\mathcal{S}'$$ It is nonempty: $$\mathcal{S}_0=(0):\quad\{\mathcal{S}_0\}\in\Lambda$$ And admits upper bounds: ...


2

Yes, you are correct. For reference: The Spectral Theorem For a Pair of Commuting Operators http://www.mi.ras.ru/~snovikov/78.pdf


0

the matters are not o straightforward. Please consult R. K. GOODRICH, The spectral theorem for real Hilbert space, Acta Sci. Math. (Szeged), 33 (1972), 123–127. (open access) Miroslav


2

As mentioned in comments, there are different "Riesz representation theorems". The Hilbert space version (pretty easy to prove) gives a natural correspondence between bounded linear functionals and the elements of the Hilbert space. The representation of linear functionals on the spaces $C(K)$ (continuous functions on a compact space), $C_c(X)$ ...


1

Because $\|y_n -x\|^2 \to d$, for all $\epsilon$ there exists $N$ s.t. $$\|y_n - x\|^2 \leq d + \epsilon$$ for all $n \geq N$. This means the last equation satisfies $$-4d + 2(\|y_n-x \|^2+ \|y_m-x\|^2) \leq -4d + 2\left((d + \epsilon) + (d + \epsilon)\right) = 4\epsilon$$ for all $m,n \geq N$. So the limit does go to zero.


0

Regard the sets: $$\Omega_n:=\{|f|\leq n\}\in\mathcal{B}(\mathbb{C})$$ Denote for shorthand: $$1_n:=\chi_{\Omega_n}\quad f_n:=f1_n$$ By dominated convergence: $$\langle f(E)\varphi,\psi\rangle=\lim_n\langle f_n(E)\varphi,\psi\rangle=\lim_n\langle\varphi,\overline{f_n}(E)\psi\rangle=\langle\varphi,\overline{f}(E)\psi\rangle$$ They extend by: ...


0

Remind the domain: $$\mathcal{D}f(E)g(E)=\mathcal{D}(fg)(E)\cap\mathcal{D}g(E)$$ One has the bound: $$\int|f|^2\mathrm{d}\nu_\varphi\leq\int(1+|f|^4)\mathrm{d}\nu_\varphi=\int|f|^4\mathrm{d}\nu_\varphi+\|\varphi\|^2$$ So for the domain: $$\mathcal{D}|f|^2\subseteq\mathcal{D}f(E)=\mathcal{D}\overline{f}(E)$$ Concluding the assertion.


1

I didn't say "at least c", btw. Say $card(S)=c$. Then $S$ has exactly $c$ countable subsets: $c$ choices for the first element, $c$ choices for the second, etc, for a total of $c^{\aleph_0}=c$. Given a countable subset, how many ways are there to assign numbers to make an element of $\ell^2(S)$? There are $c$ choices for the first coordinate, etc, again it ...


1

Suppose one has: $$\vartheta:=\|f(E)\|<\infty$$ Regard the sets: $$\Omega_n:=\{\vartheta+\frac{1}{n}\leq|f|\leq\vartheta+n\}\in\mathcal{B}(\mathbb{C})$$ And their union: $$\Omega:=\bigcup_{n=1}^\infty\Omega_n=\{|f|>\vartheta\}\in\mathcal{B}(\mathbb{C})$$ Regard the functions: $$1_n:=\chi_{\Omega_n}:\quad 1_n(E)=E(\Omega_n)$$ Denote for readability: ...


1

Meanwhile I got it... Support For the spectral radius: $$\|Z\|\leq1\implies r(Z)\leq1$$ So the support lies in: $$\mathrm{supp}E=\sigma(Z)\subseteq\overline{\mathbb{D}}$$ By the previous threads: $$g(E_Z)=\sqrt{1-Z^*Z}=\sqrt{Q}$$ So the boundary is trivial: $$\mathcal{N}g(E_Z)=(0)\implies E_Z\{g=0\}=(0)$$ (That allows to continue!) Preoperator By ...


0

By the calculus: $$f(E)E\{f=0\}=(f1_{\{f=0\}})(E)=0$$ Thus one has: $$\varphi\in\mathcal{R}E\{f=0\}\implies f(E)\varphi=0$$ Conversely regard: $$f^{-1}(\lambda):=\frac{1}{f(\lambda)}\quad(E(\lambda)\neq0)$$ By the calculus: $$f^{-1}(E)f(E)\subseteq(f^{-1}f)(E)=1(E)=1$$ But the domains agree: ...


0

Regard dense elements: $$\overline{\mathcal{D}N^*}=\mathcal{H}:\quad\varphi\in\mathcal{D}N^*$$ A calculation gives: $$ZZ^*\varphi=NQN^*\varphi=NN^*Q\varphi$$ The operator is closed: $$Q\in\mathcal{B}(\mathcal{H}):\quad NN^*=\overline{NN^*}\implies NN^*Q=\overline{NN^*Q}$$ By closed graph theorem: $$\mathcal{D}(NN^*Q)=\mathcal{H}\implies ...


1

If $i$ is an embedding, we can consider $y=i(x)$ as $x$. This just means "if we relabel $y$ and call it $x$ instead, we actually do have $X\subset Y$". Of course, it still has the usual properties of $X$ as well $y_1=i(x_1),y_2=i(x_2)\implies y_1+y_2=i(x_1+x_2)$ and similarly for scalar multiplication. The reason why this is valid is because $i(x)=i(x')$ ...


2

Denote $[T]=(a_{ij})$ and $[T^*]=(b_{ij})$. Denote further $X=(x_{ij})$ with $x_{ij}=\langle e_i,e_j\rangle$ and $Y=(y_{ij})$ with $y_{ij}=\langle b_i,b_j\rangle$. Then $ T(e_j)= \sum_k a_{kj} b_k $ and $$ \langle Te_j,b_i\rangle = \sum_k\langle a_{kj} b_k,b_i\rangle = \sum_k a_{kj} y_{ki} . $$ Analogously $T^*(b_i) = \sum_k b_{ki} e_k$, and $$ \langle ...


2

No. Let $$ S=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ T=\begin{bmatrix}0&0\\1&0\end{bmatrix}. $$ Then $$ \text{Tr}(TS)=1,\ \ \text{ and } \|T\|\,|\text{Tr}(S)|=0. $$


0

A projection $P$ should be thought of as being onto a subspace $A$ relative to a subspace $B$. As a simple example, look at $\mathbb{R}^{2}$ with standard basis $\{ e_1, e_2\}$. Another basis for $\mathbb{R}^{2}$ is $\{ e_1, e_1+e_2 \}$. The orthogonal projection of $\mathbb{R}^{2}$ onto $[\{e_1\}]$ is $Px = (x,e_1)e_1$. However, if you want to write $x$ in ...


2

Your argument is wrong because I guess you have in mind a orthogonal projector, namely a map $P: X \to X$ such that $P^2 = P$ and also $P^{\top} = P$ (or $P^* = P$). Here is a simple example. Decompose $\mathbb{R}^2 = \mathbb{R}e_1 \oplus \mathbb{R}(e_1 + e_2)$ where $e_1,e_2$ is the canonical basis i.e. $e_1=(1,0), e_2=(0,1)$. Then you have a projection $P: ...


1

You state that $X = \ker(P) \oplus \ker(P)^\perp $ and $X = \ker(P) \oplus \text{ran}(P)$ ; but then, your implication (3) that "therefore $\ker(P)^\perp = \text{ran}(P)$" rests on the assumption that the supplement space of $\ker(P)$ is unique, which is not true in general.


2

Careful, $E=F\oplus G=F \oplus H$ does not imply $G=H$.


2

For each finite subset $F \subset\subset \xi$, define $$ P_F(x) := \sum_{e \in F} \langle x,e\rangle e $$ By what you are given, the net $\{P_F\}$ converges strongly to the identity. Let $S = \overline{T(B)}$ denote the closure of the image of the unit ball under $T$, and note that $S$ is compact. For each $y \in S, \exists F_y \subset\subset \xi$ such that ...


0

I can solve this in the case that $\xi$ is a countable orthonormal basis. As my solution is too long for a comment, I am typing it here in hopes that it will aid you in the general case. $$ $$ To be concrete regarding our assumption that $\xi$ is a countable orthonormal basis, write $\xi = \{e_n\}_{n \in \mathbb{N}}$. Given $\varepsilon > 0$, we must ...



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